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Atomic term symbols

  1. Mar 22, 2008 #1
    Can someone explain to me why different terms arising from a particular electron configuration have different energies? For example, for carbon in the configuration [tex]1s^22s^22p^2[/tex] three terms arise, 3P 1D and 1S. These three terms have different energies but I don't understand why. The electron configuration is given, so what changes between the three terms? How can they have different energies?
  2. jcsd
  3. Mar 22, 2008 #2
    There are 2 electrons occupying the p-orbital. There are only 3 spin configurations the electrons can have.

    spin up-spin up
    spin down-spin down
    spin up-spin down

    These configurations are different in energy.
  4. Mar 22, 2008 #3
    Thanks for the reply! But, in an f-shell with 3 electrons for instance, even if we assume that the 3 electrons all have spin up we still get several atomic terms. How can that be, what degrees of freedom are left when the configuration is locked and the spins of the electron is also locked. The distribution of electrons within the f-shell maybe? But the [tex]m_l[/tex] quantum number of the electrons in the shells shouldn't matter for the energy should it?
  5. Mar 24, 2008 #4
    I'm not entirely sure from your post, but I think you're talking about the energy levels in the LS or Russell-Saunders coupling scheme for multi-electron atoms. The basic idea is that apart from the electrostatic attraction from the nucleus and repulsion of electrons from each other, the next greatest contributing factor is the relativistic effect of spin-orbit coupling. Loosely speaking, an electric field in one frame turns into something involving a magnetic field in another frame --- the moving electron actually sees a magnetic field. This magnetic field interacts the spin magnetic moment of the electron, to produce a further splitting between certain orbitals --- the fine structure. The term is proportional to [tex]\hat{L}\cdot\hat{S}[/tex] so [tex]\hat{J}[/tex] commutes with the Hamiltonian, where the L, S and J refer to the total angular momentum of the respective type, summed over all the electrons. Thus we can label the atomic (i.e. multi-electron states) with the term symbols [tex]^{2S+1}L_J[/tex]. Thus we need to calculate all the allowed states of S and L for a given set of electrons. It's slightly simplified as we only need to think about the electrons not in a full subshell. However, it's slightly complicated by the need to consider exchange antisymmetry.

    So the answer to your question is that different alignments of spin and angular momentum give different energies.

    If you want, I can work through some examples of calculating the allowed states, their term symbols and their relative energies (qualitatively).
  6. Mar 24, 2008 #5
    That makes sense to me... Thank you. :) And yes, I am talking about energy levels for multielectron atoms in the Russel-Saunder coupling scheme.

    But, what I am basically trying to do is to calculate the energy of excited states of some atoms without the effect of spin-orbit. Now, lets say I have an atom with the configuration [tex][Ar]3d^2[/tex] so that the two outermost electrons are d electrons and the only shell which is not full is this d shell. If the two electrons are assumed to be spin-up electrons, I get two different terms, a 3F term and a 3P term. Now, there's some energy difference between these two terms right? I just don't understand why, because the electron configuration is given and I assumed that the electrons are both spin-up electrons. So how can these two terms differ in energy? Maybe I've misunderstood something.
  7. Mar 24, 2008 #6
    Without spin-orbit coupling, single electron states with the same n number have the same energy, and multi-electron systems just have the sum of the energies. So yes, you're absolutely correct --- the energies would be the same. No amount of changing spins or orbital angular momentum orientation (m_l values) would change the energy.
  8. May 2, 2008 #7
    genneth i would love to see your worked examples, especially concerning
    L = l1+ l2, l1+ l2 -1, l1+ l2 -2,… |l1- l2|.
  9. May 2, 2008 #8
    [tex]L = l_1+ l_2, l_1+ l_2 -1, l_1+ l_2 -2,… |l_1- l_2|. [/tex]
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